Let's Make A Big Polynomial

Algebra Level 4

If $f(x)$ is an $11^\text{th}$ degree polynomial with leading coefficient $\dfrac1{2016}$ such that $f(k) = k$ for integer $1\leq k\leq 11$ , find $f(12)$ .

The answer is 19812.

1 solution

Rishabh Deep Singh
Jan 23, 2016

$1,2,3,4....11$ are the roots of the equation $f\left( x \right) -x=\frac { \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) \left( x-4 \right) .......\left( x-11 \right) }{ 2016 }$ so $f\left( 12 \right) -12=\frac { \left( 11 \right) \left( 10 \right) \left( 9 \right) \left( 8 \right) ...\left( 1 \right) }{ 2016 } =19800\\ \quad \quad \quad \quad f\left( 12 \right) =19812$

Moderator note:

Good application of the remainder factor theorem.

11!/2016+12

Dynasties Legacy - 5 years, 4 months ago

Let's Make A Big Polynomial

The answer is 19812.

1 solution

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